# Diode Logic

Discussion in 'Homework Help' started by chrispo86, Feb 18, 2009.

1. ### chrispo86 Thread Starter New Member

Jan 28, 2009
8
0
I have (what I presume is) a pretty basic question. I'm working on the attached problem and I know that when Vin is -5V, Db is ON and therefore Vx is -4.4V and Dx is off, so the output Vout is -2V. I also know that when Vin is 0V, Db is still ON and Vx is now 0.6V and Dx is now ON and Vout is 0V.

My question is about that point in between where the diode Dx turns ON. The diode should turn ON when Vx becomes greater than -2V, because there will be a more positive charge on the anode of the diode. So this will occur when Vin = -2.6V (which makes Vx = -2V). My question then is what is Vout at this point? There would still be a 0.6V drop across the diode, making Vout = -2.6V, correct? Would the graph of Vout vs Vin which the problem is looking for be what I have attached??

On a side note: is there a way to simulate an ideal diode in Microcap 9?

Thanks as always!

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2. ### Alexei Smirnov Active Member

Jan 7, 2009
43
1
You forgot about 1.067 mA current through resistors and Dx. It adds additional potential to Dx cathode. Dx goes ON when Vin is
-2 + 1.067 = -0.93 V

3. ### chrispo86 Thread Starter New Member

Jan 28, 2009
8
0
Maybe it's just been a really long day (scratch that, it's definately been a really long day), but how did you come up with 1.067 mA? And then why is that current being added to a voltage to get a new voltage?

I appologize if this is something that's painfully obvious and I'm just completely missing, my brain is pretty fried right now.

4. ### Alexei Smirnov Active Member

Jan 7, 2009
43
1
Yes, it's a really long day I was wrong.

So, starting from Vin=-5V, Db is ON, Dx is OFF, Vout=-2V

at Vin=-2V, Vx=-1.4V, Dx starts turning On, Vout=Vin. Part of the current from +5 goes to Db, part to Dx. (I hope there is no load at Vout pin?)

at Vin=-0.93, Vx=-0.33, Vout=-0.93 : all the current goes through Dx, Db turns OFF

at Vin>-0.93, Vx=-0.33, Vout=-0.93 : Dx is ON.

Looks better?

Last edited: Feb 19, 2009
5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
What is that little hiccup to -2.6V? That doesn't happen. The output is a monotonic function of the input.

6. ### chrispo86 Thread Starter New Member

Jan 28, 2009
8
0
Okay, this all makes sense to me except one thing. Vin = -0.93V. There is no load on Vout. I don't see where this number comes from. Also, why would Db turn off at this point? Vx has a higher potential than Vin, so the diode should still be on, shouldn't it?

Thanks for all the help btw, I really appreciate it.

EDIT: Okay, figured out where -0.93V came from. What i still don't see is what would Vout be when Vin = -2.4 (for example)? Vx would be -1.8V, so wouldn't Dx turn on?

I don't know why I'm having such trouble with this. I feel like this is much easier than I'm making it...

Last edited: Feb 19, 2009
7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
-0.93V is the correct upper limit. Think about this: Temporarily remove DB. Calculate Vx. You should get -0.333V. Now reconnect DB. Regardless of what happens at Vin, Vx will never go above -0.333V, because DB will be reverse biased. Since Vout is 0.6V more negative than Vx, it will never rise above -0.933V.